Mike, thanks for thinking like I had a analytic insight, but I made a
mistake.
In my defense, I did know to divide by 2, but at the time I was exploring
what forms of verbs could automatically be inverted (obverted?) to create a
possible "is_pentagonal" predicate and accidentally copied my truncated
verb into my question.
I never actually used such a form as I merely did what others did and
checked membership against the pre-generated list.
My final answer was an ugly overly-tacit overly-commutative form that
doesn't bother with tabling the negative forms (there is a high probability
that it is over-engineered, and I always strive after getting the answer to
tersifying my solution to unlock insight).
pentagonal =: -:@( [ * 1 -~ 3 * ])
(|@-)/@(0&{)@(a {~ (b {~ I.@(a e.~ (|@:-)/"1@:(a {~ b=:4&.@.@(a e.~ (
</~@:}.@i. * +/~@(a=.}.@pentagonal@i.))))))) 20000
Like I said, I know I can improve it, and was excited to see Oleg's answer
(as I knew I would) when I unlocked the forum.
Thanks for all the feedback. I'm not sure I understand Louis's feedback
yet, (seeing the anti-base there is confusing) but I am going to try all of
them now.
On Fri, Jul 12, 2019 at 9:43 AM 'Mike Day' via Programming <
[email protected]> wrote:
> It looks as if Daniel was solving the problem for pentagonal numbers * 2.
> Clearly, if there’s a solution in true Pn for a pair of indices, m > n,
> ie, (-/ Pn m,n) = +/ Pn m,n,
> it will also be a solution for 2 * Pn .
>
> I suppose there might be some half-integral solutions for 2 * Pn which
> don’t work for Pn
>
> Mike
>
> Sent from my iPad
>
> > On 12 Jul 2019, at 12:17, 'Pascal Jasmin' via Programming <
> [email protected]> wrote:
> >
> > your definition of pentagonal doesn't do the divide by 2 step
> > pentagonal =: [ -:@* ( 1 -~ 3 * ])
> >
> > this combines (,:) the +/ and -/ tables into 2, then finds intersection
> (*./). 4$.$. returns index(es)
> >
> > 4 $. $. *./@:(e.~ -/~ ,: +/~) pentagonal >: i.20115
> > 2166 1019
> >
> > On Thursday, July 11, 2019, 08:11:41 p.m. EDT, Daniel Eklund <
> [email protected]> wrote:
> >
> > Double thanks.
> >
> > One for that sparse array trick. That will go into my toolbox.
> >
> > And secondly, for the work you've been putting into the youtube videos
> -- I
> > think they've been an invaluable teaching aid, and a welcome addition to
> > the relative paucity of J learning outside of the main website.
> >
> >
> > On Thu, Jul 11, 2019 at 7:35 PM 'robert therriault' via Programming <
> > [email protected]> wrote:
> >
> >> Wow Daniel,
> >> I am sincerely impressed at how you wrestled that one to the ground.
> >>
> >> A trick I learned a while ago on these forums was the use of Sparse ($.)
> >>
> >> toy e. _1
> >> 1 0 0 0
> >> 0 0 1 1
> >> 0 0 0 0
> >> $. toy e. _1 NB. converts dense array to sparse form
> >> 0 0 │ 1
> >> 1 2 │ 1
> >> 1 3 │ 1
> >> 4 $. $. toy e. _1 NB. Dyadic 4 $. returns the indices of sparse form
> >> 0 0
> >> 1 2
> >> 1 3
> >>
> >> Cheers, bob
> >>
> >>> On Jul 11, 2019, at 4:20 PM, Daniel Eklund <[email protected]> wrote:
> >>>
> >>> Hi everyone,
> >>>
> >>> I’m looking for some newbie help. I feel I’ve come so far but I’ve run
> >>> into something that is making me think I’m not really getting something
> >>> fundamental.
> >>>
> >>> Rather than try to come up with a contrived example, I’ll just say
> >> outright
> >>> that I’m trying to solve one of the project Euler questions (problem
> 44)
> >>> and in my desire to use a particular J strategy (‘tabling’) I’m
> >> struggling
> >>> to deduce how array indexing to recover the input pairs works best.
> >>>
> >>> From the question:
> >>>
> >>> A pentagonal number is Pn=n(3n−1)/2. The first ten pentagonal numbers
> >> are:
> >>>
> >>> 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, …
> >>>
> >>> The challenge is to find two pentagonal numbers that both added
> together
> >>> and whose difference is also a pentagonal number.
> >>>
> >>> Producing pentagonals is easy enough:
> >>>
> >>> pentagonal =: [ * ( 1 -~ 3 * ])
> >>>
> >>> pentagonal >: i. 5
> >>>
> >>> 2 10 24 44 70
> >>>
> >>> My plan was then to table both the addition and subtraction
> >>>
> >>> +/~ pentagonal >: i. 5
> >>>
> >>> 4 12 26 46 72
> >>>
> >>> 12 20 34 54 80
> >>>
> >>> 26 34 48 68 94
> >>>
> >>> 46 54 68 88 114
> >>>
> >>> 72 80 94 114 140
> >>>
> >>> -/~ pentagonal >: i. 5
> >>>
> >>> 0 _8 _22 _42 _68
> >>>
> >>> 8 0 _14 _34 _60
> >>>
> >>> 22 14 0 _20 _46
> >>>
> >>> 42 34 20 0 _26
> >>>
> >>> 68 60 46 26 0
> >>>
> >>> And then fetch the values in the table that were also pentagonal. This
> >>> seems like a sane strategy -- not entirely clever -- but I am running
> >> into
> >>> something that makes me feel I am missing something essential: how to
> >>> recover the numbers (horizontal row number and column number pair, and
> >>> therefore input values) that induced the result.
> >>>
> >>> I create myself a toy example to try to understand, by creating a
> >> hardcoded
> >>> matrix where I’m interested in those that have a certain value:
> >>>
> >>> ] toy =: 3 4 $ _1 2 12 9 32 23 _1 NB. _1 will be value of
> interest
> >>>
> >>> _1 2 12 9
> >>>
> >>> 32 23 _1 _1
> >>>
> >>> 2 12 9 32
> >>>
> >>> toy e. _1 NB. I am searching for _1 (a proxy for a truth function
> like
> >>> "is_pentagonal")
> >>>
> >>> 1 0 0 0
> >>>
> >>> 0 0 1 1
> >>>
> >>> 0 0 0 0
> >>>
> >>> I cannot easily find the indices using I., because
> >>>
> >>> I. toy e. _1
> >>>
> >>> 0 0
> >>>
> >>> 2 3
> >>>
> >>> 0 0
> >>>
> >>> Has a zero in the first row which is semantically important, but
> >>> zero-padding (semantically unimportant) in other locations.
> >>>
> >>> I realize I can ravel the answer and deduce (using the original shape)
> >> the
> >>> indices
> >>>
> >>> I. , toy e. _1
> >>>
> >>> 0 6 7
> >>>
> >>> NB. The following is ugly, but works
> >>>
> >>> (<.@:%&4 ; 4&|)"0 I. , toy e. _1 NB. The 4 hardcoded is the
> length
> >>> of each item
> >>>
> >>> ┌─┬─┐
> >>>
> >>> │0│0│
> >>>
> >>> ├─┼─┤
> >>>
> >>> │1│2│
> >>>
> >>> ├─┼─┤
> >>>
> >>> │1│3│
> >>>
> >>> └─┴─┘
> >>>
> >>> And voilà a table of items of row/column pairs that can be used to
> fetch
> >>> the original inducing values.
> >>>
> >>> To get to the point: is there anything easier? I spent a long time on
> >>> NuVoc looking at the “i.” family along with “{“. I feel like I might
> be
> >>> missing out on something obvious, stipulating that there is probably
> >>> another way to do this without tabling and trying to recover the
> inducing
> >>> values.
> >>>
> >>> Is my desire, i.e. to table results and simultaneously keep pointers
> back
> >>> to the original values (a matter of some hoop jumping) the smell of an
> >>> anti-pattern in J ?
> >>>
> >>> Thanks for any input.
> >>>
> >>> Daniel
> >>>
> >>> PS. In my question I purposefully am ignoring the obvious symmetry of
> >>>
> >>> +/~ pentagonal >: i. 5
> >>>
> >>> As I am interested in the _general_ case of tabling two different
> arrays:
> >>>
> >>> ( pentagonal 20+i.6) +/ pentagonal >: i. 5
> >>>
> >>> 1182 1190 1204 1224 1250
> >>>
> >>> 1304 1312 1326 1346 1372
> >>>
> >>> 1432 1440 1454 1474 1500
> >>>
> >>> 1566 1574 1588 1608 1634
> >>>
> >>> 1706 1714 1728 1748 1774
> >>>
> >>> 1852 1860 1874 1894 1920
> >>>
> >>> And recovering their indices.
> >>> ----------------------------------------------------------------------
> >>> For information about J forums see http://www.jsoftware.com/forums.htm
> >>
> >> ----------------------------------------------------------------------
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> >>
> > ----------------------------------------------------------------------
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> >
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>
> ----------------------------------------------------------------------
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